質問編集履歴

3

2019/12/20 01:20

投稿

m9288650228
m9288650228

スコア18

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+ ### 参考程度のSQLです。
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  ```SQL
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  SELECT kind_id, COUNT(*) FROM sample WHERE fruits_id IN (1, 2) GROUP BY kind_id HAVING COUNT(*) = 2;

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kind_id

2019/12/20 01:19

投稿

m9288650228
m9288650228

スコア18

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  ```SQL
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- SELECT col1, COUNT(*) FROM table1 WHERE col2 IN (1, 2) GROUP BY col1 HAVING COUNT(*) = 2;
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+ SELECT kind_id, COUNT(*) FROM sample WHERE fruits_id IN (1, 2) GROUP BY kind_id HAVING COUNT(*) = 2;
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  ```
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1

2019/12/20 01:03

投稿

m9288650228
m9288650228

スコア18

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  ```
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+ ```SQL
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+ SELECT col1, COUNT(*) FROM table1 WHERE col2 IN (1, 2) GROUP BY col1 HAVING COUNT(*) = 2;
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+ ```
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+ 上記も参考にしましたがこれでは動的にならないのでできませんでした。